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In my calculus textbook there is a part explaining tangent planes to surfaces, and how to form the normal vector for such a plane. I get the idea (taking the cross product of the tangential vectors) but I'm sort of confused regarding the way there.

We start by forming a tangent plane, as well as the vertical planes $x=a$ and $y=b$. The tangent plane then intersects these vertical planes in straight lines, each with slope $f_1(a,b)$ and $f_2(a,b)$ (partial derivatives). This I am 100% onboard with. Then they go on to say that these vectors are parallel to the vectors $T_1 = \mathbf{i} + f_1(a,b) \mathbf{k}$ and $T_2 = \mathbf{j} + f_2(a,b) \mathbf{k}$ respectively. This is where I get lost, in that I dont really see what the notation here represents. To me it looks like a position vector in the $x,z$ plane of the form $r = \langle 1, f_1(a,b)\rangle$ but that doesn't make any sense. Also it doesnt match up to the picture they have with $T_1$ lining up perfectly along the tangent plane (which is what I expected it to do!).

So yeah, how do you interpret these vectors? They look like position vectors in notation but obviously aren't, as they go along the plane. It's like I understand what they are saying but not their way of writing it! I am used to seeing vectors in these contexts as either position vectors or as the difference between two points on the plane.

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A vector has no inherent position in space. It only represents a magnitude and a direction. So we might as well imagine that $T_1$ and $T_2$ are based at $(a,b).$ (If you want, think of it as a new coordinate system where $(a,b)$ is now the origin. Or, going the other way around, translate the point $(a,b)$ to the origin, so that we may now cast everything in terms of linear algebra.) In this framework all they're saying is that the vector $T_1$ spans the line with slope $f_1(a,b)$ and the vector $T_2$ spans the line with slope $f_2(a,b).$

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  • $\begingroup$ Hmm! I get what you are saying, but I am still having a bit of a hard time "seeing it". Am I correct in that the vector $T_1 = \mathbf{i} + f_1(a,b) \mathbf{k}$ is the same thing as writing $<1,y,f1(a,b)>$? If so, then how does that describe anything like a tangent vector to the surface, it would seem to just be a straight line parallel to the y-axis? (I am assuming that I am very wrong about this, but I can't really see where) $\endgroup$ – Laplacinator Jun 23 '15 at 18:00
  • $\begingroup$ In your notation, you would write the vector $T_1$ as $\langle 1,0,f_1(a,b) \rangle,$ since there is no $y$-component in $T_1.$ Imagine that you move your surface so that the point $(a,b)$ is exactly at the origin of your coordinate system. Then is it easier to see how $T_1$ is a vector that spans the line which is the intersection of the tangent plane to the surface with the plane $y=0$ (i.e. the $xz$-plane)? $\endgroup$ – Alex Provost Jun 23 '15 at 19:45

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